Question

Find the probability of at most two tails or at least two heads in a toss of three coins.

Hint: Here we proceed the solution by considering all possible outcomes of three coins when they are tossed. So here we have to find the probability of at most two tails or at least two heads where at most indicates two or less than two.

Let set S be the sample space of all three possible outcomes of three coins tossed.
S= {HHH, HTH, THH, HHT, TTH, HTT, THT, TTT}
Here H denotes Heads and T denotes Tail
Now from the set S
The total number of outcomes = 8
Here we have to find probability of at most two tails or at least two heads
As we know that at most 2 means two or less than two. So now by using the at most condition, we can say
Number of outcomes with at-most two tails or at least two heads = 7
Now by using the total number of outcomes and number of outcomes with at-most two tails or at-least two heads we can find the required probability.

Therefore probability of at-most two tails and at least two in a toss of three coins =
$= \dfrac{{{\text{Number of outcomes with at most 2tails or at least 2heads}}}}{{{\text{total number of outcomes}}}} \\ = \dfrac{7}{8} \\$

Note: In this problem we have to focus on two words they are at-most and at-least these are the two key words the whole problem depends on these are conditions these are conditions given to solve the problem. Generally everyone will ignore these words which will affect the answer.